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3 years ago

Una tortuga recorre una pista de 1,25 km que tiempo emplearia si lleara una velocidad de 0,2 km/h

Physics

1 answer:

Marina86 [1]3 years ago

7 0

Responder:

<h2>5 horas </h2>

Explicación:

La velocidad se define como el cambio de distancia de un cuerpo con respecto al tiempo. Matemáticamente, Velocidad = Distancia / Tiempo

Dada la velocidad del coche = 0,25 km / hy la distancia = 1,25 km

El tiempo se expresa a partir de la fórmula como Tiempo = Distancia / Velocidad

La sustitución de los valores dados en la fórmula dará;

Tiempo = 1,25 km / 0,25 km / h

Tiempo = 5 horas

Por lo tanto, la tortuga tardará 5 horas en viajar a una velocidad de 0,2 km / h.

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2 years ago

The 5-kg block A has an initial speed of 5 m/s as it slides down the smooth ramp, after which it collides with the stationary bl

Akimi4 [234]

Answer:

The coupled velocity of both the blocks is 1.92 m/s.

Explanation:

Given that,

Mass of block A, $m_1=5\ kg$

Initial speed of block A, $u_1=5\ m/s$

Mass of block B, $m_2=8\ kg$

Initial speed of block B, $u_2=0$

It is mentioned that if the two blocks couple together after collision. We need to find the common velocity immediately after collision. We know that due to coupling, it becomes the case of inelastic collision. Using the conservation of linear momentum. Let V is the coupled velocity of both the blocks. So,

$m_1u_1+m_2u_2=(m_1+m_2)V\\\\V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\V=\dfrac{5\times 5+0}{(5+8)}\\\\V=1.92\ m/s$

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3 years ago

Find the object's speeds v1, v2, and v3 at times t1=2.0s, t2=4.0s, and t3=13s.

Burka [1]

Since this is a distance/time graph, the speed at any time is the slope
of the part of the graph that's directly over that time on the x-axis.

At time t1 = 2.0 s
That's in the middle of the first segment of the graph,
that extends from zero to 3 seconds.
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At time t2 = 4.0 s
That's in the middle of the horizontal part of the graph
that runs from 3 to 6 seconds.
Its slope is zero.
v2 = zero .

At time t3 = 13 s.
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3 years ago

Read 2 more answers

0/2 File Limit

slamgirl [31]

Answer:

Speed at which it will reach the ground is given as

$v_f = 46.8 m/s$

Total time for which it will remain in air is given as

t = 6.3 s

Explanation:

As we know that the object is projected upwards with speed

$v_i = 15 m/s$

$g = - 9.81 m/s^2$

now when it will reach the ground then we have

$y = v_y t + \frac{1}{2} at^2$

so we have

$-100 = 15 t - \frac{1}{2}(-9.81) t^2$

$4.905 t^2 - 15 t - 100 = 0$

so we have

$t = 6.3 s$

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$v_f = v_i + at$

$v_f = -15 + (9.81)(6.3)$

$v_f = 46.8 m/s$

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3 years ago

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3 years ago